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To determine whether the given series converges absolutely or diverges, we'll apply the **Ratio Test**. The Ratio Test states that for a series \( \sum a_n \), if 

\[
L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|
\]

then:
1. If \( L < 1 \), the series converges absolutely.
2. If \( L > 1 \), the series diverges.
3. If \( L = 1 \), the test is inconclusive.

### Given Series
The series is:

\[
\sum_{n=1}^{\infty} \frac{(-1)^n n^2 (n+4)!}{n! 9^{3n}}
\]

This gives us the general term:

\[
a_n = \frac{(-1)^n n^2 (n+4)!}{n! 9^{3n}}
\]

### Step 1: Set up \( a_{n+1} \)
To use the Ratio Test, we need \( a_{n+1} \):

\[
a_{n+1} = \frac{(-1)^{n+1} (n+1)^2 (n+5)!}{(n+1)! 9^{3(n+1)}}
\]

### Step 2: Calculate \( \left| \frac{a_{n+1}}{a_n} \right| \)
Now, let’s compute the ratio \( \left| \frac{a_{n+1}}{a_n} \right| \):

\[
\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(-1)^{n+1} (n+1)^2 (n+5)!}{(n+1)! 9^{3(n+1)}} \cdot \frac{n! 9^{3n}}{(-1)^n n^2 (n+4)!} \right|
\]

Simplifying this expression:

1. **Cancel out \( (-1)^{n+1} \) with \( (-1)^n \)**: The absolute value will eliminate any negative signs.
2. **Simplify the factorials**: \((n+5)! / (n+4)! = (n+5)\).
3. **Simplify the powers of 9**: \(9^{3n} / 9^{3(n+1)} = \frac{1}{9^3} = \frac{1}{729}\).

Thus, we get:

\[
\left| \frac{a_{n+1}}{a_n} \right| = \frac{(n+1)^2 (n+5)}{(n+1) \cdot 729 \cdot n^2}
\]

Further simplifying:

\[
= \frac{(n+1) (n+5)}{729 n^2}
\]

### Step 3: Take the Limit as \( n \to \infty \)
Now, we take the limit of \( \left| \frac{a_{n+1}}{a_n} \right| \) as \( n \to \infty \):

\[
\lim_{n \to \infty} \frac{(n+1)(n+5)}{729 n^2} = \lim_{n \to \infty} \frac{n^2 + 6n + 5}{729 n^2} = \frac{1}{729}
\]

Since \( \frac{1}{729} < 1 \), the Ratio Test tells us that the series **converges absolutely**.

### Conclusion
The series converges absolutely by the Ratio Test.

Would you like further clarification or details on any step?

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Here are some related questions to deepen your understanding:

1. What is absolute convergence, and how does it differ from conditional convergence?
2. Can the Ratio Test be applied to any series, or are there specific conditions?
3. Why does the presence of \( (-1)^n \) in a series affect its convergence?
4. What happens if the Ratio Test gives \( L = 1 \)?
5. How does the Ratio Test compare to other tests like the Root Test for convergence?

**Tip:** When simplifying factorials in ratios, remember that \( (n+1)! = (n+1) \cdot n! \), which helps cancel terms efficiently.

Use the Ratio Test to determine if the following series converges absolutely or diverges: ∑ ( (-1)^n * n^2 * (n+4)! ) / (n! * 9^(3n)), with n going from 1 to infinity.

Learn how to apply the Ratio Test to determine the absolute convergence or divergence of the series ∑ ( (-1)^n * n^2 * (n+4)! ) / (n! * 9^(3n)). This step-by-step solution covers key calculus concepts, including factorial simplification and limits, suitable for advanced calculus students.

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